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October 1, 2021, 07:38 
Unsteady simulation gives steady result

#1 
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Hello! To keep things short, I was doing a RANS simulation where the residuals wouldn't converge. Instead they showed the typical oscillating pattern that I would associate with some transient behavior of the solution. They did neither grow smaller but oscillated around a constant value.
So I though I'd set up a transient simulation, starting from the oscillating solution I had. I did so, and to my surprise the transient simulation gave a steady solution, i.e. after some small initial fluctuations it turned steady, and has been so ever since. At first I though my total physical time I had set was to short to show any unsteady behavior, but as I've let it simulate some more, and by looking at a for me relevant statistics/plots, I'm starting to believe it wouldn't change even if I let it run forever. Obviously, I cannot let it do that however. So is this something that can happen? I.e that a steady simulation fails to find a steady solution and one then finds it via a unsteady simulation? I'm a beginner in CFD and have limited experience in unsteady simulations. 

October 1, 2021, 09:15 

#2 
Senior Member

We can't obviously give an answer that is specific for your case without knowing each detail but, in general, this is certainly a possibility.
Very roughly speaking, to understand why, you need to know that most (if not all) the algorithms for steady cases are, indeed, in a form that can be related to a sort of unsteady algorithm. The difference between this said unsteady algorithm and the steady one is then only in the local time steps used for each computational cell. The unsteady algorithm will have a single, unique time step for the whole domain, and it is (hopefully) respectful of the accuracy and stability required by the scheme and problem. The steady algorithm won't have any accuracy limitation, and will thus force the local time step to be the highest possible (according to stability) to get to the steady state faster. Now, I should double check, but I'm pretty sure that, for linear problems, this difference between steady and unsteady algorithms can't, as a matter of fact, produce the disalignment you have experienced between the two solutions. No matter what (but within the stability limits of the algorithm). However, for non linear problems, the necessary linearization to advance the solution toward steady state might be along a direction and of an amount which is getting you into a limit cycle. In contrast, the unsteady algorithm would have, for the same initial condition, much different direction and amount of the advancement. Notably, the unsteady algorithm always moves slower than the steady one. A rough example of this would be driving a car looking for a parking lot and actually parking the car. If you keep going at, say, 100 km/h, you won't spot any parking lot at all and won't be able to park, ever. A possible way to test this would be reducing the under relaxation factors of the steady algorithm, that are always related to the local time step in a proportional way. However, let me say that, to me, getting to the steady state with an unsteady algorithm, provided that everything is made in the correct manner, is actually more solid than getting there with a steady algorithm, so I would be perfectly fine with the steady solution you got with the unsteady algorithm. Depending from how much into CFD you are, you might find helpful the following material on the topic: https://www.researchgate.net/publica...ime_Derivative 

October 1, 2021, 12:30 

#3 
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Filippo Maria Denaro
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I agree that if you know you have astatistically steady problem, the URANS is correctly answering to your problem providing a steady solution.
On the other hand, we should take care of what is your definition of "steady" solution when running the URANS. How do you decide that the solution is steady? Are you sure that you have the time derivatives really going to zero and not a case where they start oscillating around a small but constant value? That would be as same as you get in the steady case. 

October 1, 2021, 13:44 

#4  
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rans model, steady and unsteady state 
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